Math, asked by binodkumaryadav615, 7 months ago

Prove that √7
is an errational no.​

Answers

Answered by Anonymous
2

Answer:

hey dude here u can take7 instead of 2 okk

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Answered by CEOSanjay
0

Answer :

let us assume that √7 be rational.

then it must in the form of p / q.

As definition of rational number says.. P is whole number q is non zero whole number.. And p and q is simplest ratio which is expressed.. That means there exists no prime factor common in p and q.

√7 = p / q

√7 x q = p

squaring on both sides

7q² = p² ------1.

p is divisible by 7

p = 7c [c is a positive integer] [squaring on both sides ]

p²= 49c²

subsitute p² in eqn(1) we get

7q² = 49 c²

q² = 7c²

q is divisble by 7

thus q and p have a common factor 7.

there is a contradiction to our assumption

as our assumsion p & q are co prime but it has a common factor.

so that √7 is an irrational.

Happy Learning..!!

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